# 2011 AMC 10B Problems/Problem 23

## Problem

What is the hundreds digit of $2011^{2011}?$

$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

## Solution 1

Since $2011 \equiv 11 \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011} \pmod{1000}.$

To compute this, we use a clever application of the binomial theorem.

\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}

In all of the other terms, the power of $10$ is greater than $3$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:

\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}

Therefore, the hundreds digit is $\boxed{\textbf{(D) } 6}.$

Sidenote: By Euler's Totient Theorem, $a^{\phi (1000)} \equiv 1 \pmod{1000}$ for any $a$, so $a^{400} \equiv 1 \pmod{1000}$ and $11^{2011} \equiv 11^{11} \pmod{1000}$. We can then proceed using the clever application of the Binomial Theorem.

## Solution 2

We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem, it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$

In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have $$2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.$$

In modulo $125,$ we have $2011^{100} \equiv 1 \pmod{125}$ by Euler's Theorem, and also $2011 \equiv 11 \pmod{125}.$ Therefore, we have \begin{aligned} 2011^{2011} &= (2011^{100})^{20} \cdot 2011^{11} \\ &\equiv 1^{20} \cdot 11^{11} \\ &= 121^5 \cdot 11 \\ &= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &\equiv -24 \cdot 11 = -264 \\ &\equiv 111 \pmod{125}. \end{aligned}

After finding the solution $2011^{2011} \equiv 611 \pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\boxed{\textbf{(D) } 6}.$

## Solution 3

Notice that the hundreds digit of $2011^{2011}$ won't be affected by $2000$. Essentially we could solve the problem by finding the hundreds digit of $11^{2011}$. Powers of $11$ are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of $11$ can be found by reading rows of the triangle and adding extra numbers up. [add source] For example, the sixth row of the triangle is $1, 5, 10, 10, 5,$ and $1$. Adding all numbers from right to left, we get $161051$, which is also $11^5$. In other words, each number is $10^n$ steps from the right side of the row. The hundreds digit is $0$. We can do the same for $11^{2011}$, but we only need to find the $3$ digits from the right. Observing, every $3$ number from the right is $1 + 2 + 3... + n$. So to find the third number from the right on the row of $11^{2011}$, $$f(11^n) = 1 + 2 + 3... + (n-1),$$ or $\frac{(2010 \cdot 2011)}{2}$, or $2021055$. The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously $2011$. We must carry the $1$ in $2011$'s tens digit to the $5$ in $2021055$'s unit digit to get $\boxed{\textbf{(D) } 6}$. The one at the very end of the row doesn't affect anything, so we can leave it alone.

-jackshi2006